Problem: $\begin{cases}b(1)=-2\\\\ b(n)=b(n-1)-7 \end{cases}$ Find the $3^{\text{rd}}$ term in the sequence.
Explanation: This is a recursive formula. It tells us that the first term is $-2$ and that the common difference is $-7$. $\begin{aligned} {b(1)}&=-2 \\\\ {b(2)}&={b(1)}-7=-9 \\\\ {b(3)}&={b(2)}-7=-16 \end{aligned}$ The $3^{\text{rd}}$ term is $-16$.